Exponential Growth of Bacteria

Date: 7 Aug 1995 08:19:31 -0400
From: Greg Sharpe
Subject: Growth & Decay
A colony of N bacteria increases with time according to the formula
dN
-- = kN where K is constant.
dT
If initially there are N bacteria show that the number at any time, N,
0
kt
is given by N = N e
0
If the number of bacteria doubles every 2 hours, how long (to the nearest
minute) will the colony take to treble?

Date: 8 Aug 1995 22:35:26 -0400
From: Dr. Ken
Subject: Re: Growth & Decay
Hello there!
This is an example of a differential equation. To solve it, we'll use
integration:
dN
-- = kN
dt
dN = kN * dt
1/N * dN = k * dT Then integrate:
log[N] = k*t + C
N = e^(k*t + C) = e^C * e^kt
To find C (or rather e^C), we plug in known simultaneous values of N and t:
N = e^C * e^0 = e^C
0
So if you plug back in, you get the desired equation.
>If the number of bacteria doubles every 2 hours, how long (to the nearest
>minute) will the colony take to treble?
The information "doubles every 2 hours" lets us actually find k. We'll plug
in k=0 and k=1.
Since we know N = N e^kt, we know 2*N = N , i.e. 2*N e^0 = N e^k, so
0 0 1 0 0
e^k = 2. Therefore k = Log[2]. Right? So then we have
N = N e^(Log[2] * t). We can find out how long it will take to treble by
0
figuring out when the amount of N is 3*N . So we have the equation
0
3N = N e^(Log[2] * t), which you can tackle with a calculator.
0 0
Hope this helps!
-K